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Let me just slightly restate your problem to make sure I'm being clear. Let's say we have frame A and frame B, where $\vec{v}_{BA} = \vec{u}$, meaning the "velocity of B with respect to A" is $\vec{u}$. Then if we have an object, C, with velocity $\vec{v}_{CA} = \vec{v}$, its velocity with respect to frame B is (using the relative velocity equation): $$ \vec{v}_{CB} = \vec{v}_{CA} + \vec{v}_{AB} = \vec{v} + \left( - \vec{v}_{BA} \right) = \vec{v} - \vec{u} \,. $$ Now you are assuming that Newton's Second Law says that the net force on an object over time can be found kinematically from the momentum: $$ F_C = \frac{d \vec{p}_C(t)}{d t} \quad \text{with} \quad \vec{p}(t) = m(t) \, \vec{v}(t)\,. $$ Then, for frame A, we have: $$ \begin{align} \vec{F}_C^{(A)} &= \frac{d \vec{p}^{(A)}_C(t)}{d t}\\ & = \frac{d m_C}{dt}\, \vec{v}_{CA} + m_C \, \frac{d \vec{v}_{CA}}{dt} = \frac{d m_C}{dt}\, \vec{v} + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ and for frame B: $$ \begin{align} \vec{F}_C^{(B)} &= \frac{d \vec{p}^{(B)}_C(t)}{d t} \\ & = \frac{d m_C}{dt}\, \vec{v}_{CB} + m_C \, \frac{d \vec{v}_{CB}}{dt} \\ & = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \left(\vec{v} - \vec{u} \right)}{dt} = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ where the last equality follows because $\vec{u}$ is a constant. So we find that: $$ \left(\vec{F}_C^{(A)} - \vec{F}_C^{(B)}\right)(t) = \frac{dm_C(t)}{dt} \, \vec{u} $$ Is that your conclusion?

I think the issue is that your statement of Newton's Second Law is incorrect. The law is the equation of motion for a particle of mass $m$. In that way we can write Newton II in all of these equivalent ways: $$ \vec{F}_\text{net} = m \vec{a} = m \frac{d\vec{v}}{dt} = \frac{d \left( m \vec{v}\right)}{dt} = \frac{d \vec{p}}{dt} $$ exactly because we assume that the particle has constant mass.

If you would like to write down a "Newton's Second Law for an object with changing mass", then you need to derive an effective equation for a system of particles. See, for example, my recent answer here. The result of summing many Newton II equations for many particles is this: $$ \vec{F}_\text{net, external} = M_\text{tot} \frac{d \vec{V}_\text{com}}{d t}\,, $$ where the left-hand side is the net external force on the system (all internal forces have cancelled out via Newton III); $M_\text{tot}$ is the sum of all the particle masses; and the center-of-mass velocity is defined as: $$ \vec{V}_\text{com} = \frac{d}{dt} \vec{R}_\text{com} = \frac{d}{dt} \left( \frac{m_1 \vec{v}_1 + \cdots + m_N \vec{v}_N}{M_\text{tot}} \right)\,. $$

I think you need to frame your question in terms of that extension of Newton's Second Law. I'm not sure how to precisely state such a question, but it would be a nice exercise. I don't think, though, that in doing so you will arrive at such contradictions, because to have a contradiction you'll need to show it happening at the level of the equation of Newton's Second Law for a particle.

In your original statement, which I summarized at the beginning of the answer, there is simply no contradiction because it is impossible, within the theory, for a particle $m$ to have changing mass.

Let me just slightly restate your problem to make sure I'm being clear. Let's say we have frame A and frame B, where $\vec{v}_{BA} = \vec{u}$, meaning the "velocity of B with respect to A" is $\vec{u}$. Then if we have an object, C, with velocity $\vec{v}_{CA} = \vec{v}$, its velocity with respect to frame B is (using the relative velocity equation): $$ \vec{v}_{CB} = \vec{v}_{CA} + \vec{v}_{AB} = \vec{v} + \left( - \vec{v}_{BA} \right) = \vec{v} - \vec{u} \,. $$ Now you are assuming that Newton's Second Law says that the net force on an object over time can be found kinematically from the momentum: $$ F_C = \frac{d \vec{p}_C(t)}{d t} \quad \text{with} \quad \vec{p}(t) = m(t) \, \vec{v}(t)\,. $$ Then, for frame A, we have: $$ \begin{align} \vec{F}_C^{(A)} &= \frac{d \vec{p}^{(A)}_C(t)}{d t}\\ & = \frac{d m_C}{dt}\, \vec{v}_{CA} + m_C \, \frac{d \vec{v}_{CA}}{dt} = \frac{d m_C}{dt}\, \vec{v} + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ and for frame B: $$ \begin{align} \vec{F}_C^{(B)} &= \frac{d \vec{p}^{(B)}_C(t)}{d t} \\ & = \frac{d m_C}{dt}\, \vec{v}_{CB} + m_C \, \frac{d \vec{v}_{CB}}{dt} \\ & = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \left(\vec{v} - \vec{u} \right)}{dt} = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ where the last equality follows because $\vec{u}$ is a constant. So we find that: $$ \left(\vec{F}_C^{(A)} - \vec{F}_C^{(B)}\right)(t) = \frac{dm_C(t)}{dt} \, \vec{u} $$ Is that your conclusion?

I think the issue is that your statement of Newton's Second Law is incorrect. The law is the equation of motion for a particle of mass $m$. In that way we can write Newton II in all of these equivalent ways: $$ \vec{F}_\text{net} = m \vec{a} = m \frac{d\vec{v}}{dt} = \frac{d \left( m \vec{v}\right)}{dt} = \frac{d \vec{p}}{dt} $$ exactly because we assume that the particle has constant mass.

If you would like to write down a "Newton's Second Law for an object with changing mass", then you need to derive an effective equation for a system of particles. See, for example, my recent answer here. The result of summing many Newton II equations for many (say, $N$) particles is this: $$ \vec{F}_\text{net, external} = M_\text{tot} \frac{d \vec{V}_\text{com}}{d t}\,, $$ where the left-hand side is the net external force on the system (all internal forces have cancelled out via Newton III); $M_\text{tot}$ is the sum of all the particle masses; and the center-of-mass velocity is defined as: $$ \vec{V}_\text{com} = \frac{d}{dt} \vec{R}_\text{com} = \frac{d}{dt} \left( \frac{m_1 \vec{r}_1 + \cdots + m_N \vec{r}_N}{M_\text{tot}} \right)\,. $$

I think you need to frame your question in terms of that extension of Newton's Second Law. I'm not sure how to precisely state such a question, but it would be a nice exercise. I don't think, though, that in doing so you will arrive at such contradictions, because to have a contradiction you'll need to show it happening at the level of the equation of Newton's Second Law for a particle.

In your original statement, which I summarized at the beginning of the answer, there is simply no contradiction because it is impossible, within the theory, for a particle $m$ to have changing mass.

Let me just slightly restate your problem to make sure I'm being clear. Let's say we have frame A and frame B, where $\vec{v}_{BA} = \vec{u}$, meaning the "velocity of B with respect to A" is $\vec{u}$. Then if we have an object, C, with velocity $\vec{v}_{CA} = \vec{v}$, its velocity with respect to frame B is (using the relative velocity equation): $$ \vec{v}_{CB} = \vec{v}_{CA} + \vec{v}_{AB} = \vec{v} + \left( - \vec{v}_{BA} \right) = \vec{v} - \vec{u} \,. $$ Now you are assuming that Newton's Second Law says that the net force on an object over time can be found kinematically from the momentum: $$ F_C = \frac{d \vec{p}_C(t)}{d t} \quad \text{with} \quad \vec{p}(t) = m(t) \, \vec{v}(t)\,. $$ Then, for frame A, we have: $$ \begin{align} \vec{F}_C^{(A)} &= \frac{d \vec{p}^{(A)}_C(t)}{d t}\\ & = \frac{d m_C}{dt}\, \vec{v}_{CA} + m_C \, \frac{d \vec{v}_{CA}}{dt} = \frac{d m_C}{dt}\, \vec{v} + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ and for frame B: $$ \begin{align} \vec{F}_C^{(B)} &= \frac{d \vec{p}^{(B)}_C(t)}{d t} \\ & = \frac{d m_C}{dt}\, \vec{v}_{CB} + m_C \, \frac{d \vec{v}_{CB}}{dt} \\ & = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \left(\vec{v} - \vec{u} \right)}{dt} = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ where the last equality follows because $\vec{u}$ is a constant. So we find that: $$ \left(\vec{F}_C^{(A)} - \vec{F}_C^{(B)}\right)(t) = \frac{dm_C(t)}{dt} \, \vec{u} $$ Is that your conclusion?

I think the issue is that your statement of Newton's Second Law is incorrect. The law is the equation of motion for a particle of mass $m$. In that way we can write Newton II in all of these equivalent ways: $$ \vec{F}_\text{net} = m \vec{a} = m \frac{d\vec{v}}{dt} = \frac{d \left( m \vec{v}\right)}{dt} = \frac{d \vec{p}}{dt} $$ exactly because we assume that the particle has constant mass.

If you would like to write down a "Newton's Second Law for an object with changing mass", then you need to derive an effective equation for a system of particles. See, for example, my recent answer here. The result of summing many Newton II equations for many particles is this: $$ \vec{F}_\text{net, external} = M_\text{tot} \frac{d \vec{V}_\text{com}}{d t} $$ where the left-hand side is the net external force on the system (all internal forces have cancelled out via Newton III), $M_\text{tot}$ is the sum of all the particle masses, and the center-of-mass velocity is defined as: $$ \vec{V}_\text{com} = \frac{d}{dt} \vec{R}_\text{com} = \frac{d}{dt} \left( \frac{m_1 \vec{v}_1 + \cdots + m_N \vec{v}_N}{M_\text{tot}} \right)\,. $$

I think you need to frame your question in terms of that extension of Newton's Second Law. I'm not sure how to precisely state such a question, but it would be a nice exercise. I don't think, though, that in doing so you will arrive at such contradictions, because to have a contradiction you'll need to show it happening at the level of the equation of Newton's Second Law for a particle.

In your original statement, which I summarized at the beginning of the answer, there is simply no contradiction because it is impossible, within the theory, for a particle $m$ to have changing mass.

Let me just slightly restate your problem to make sure I'm being clear. Let's say we have frame A and frame B, where $\vec{v}_{BA} = \vec{u}$, meaning the "velocity of B with respect to A" is $\vec{u}$. Then if we have an object, C, with velocity $\vec{v}_{CA} = \vec{v}$, its velocity with respect to frame B is (using the relative velocity equation): $$ \vec{v}_{CB} = \vec{v}_{CA} + \vec{v}_{AB} = \vec{v} + \left( - \vec{v}_{BA} \right) = \vec{v} - \vec{u} \,. $$ Now you are assuming that Newton's Second Law says that the net force on an object over time can be found kinematically from the momentum: $$ F_C = \frac{d \vec{p}_C(t)}{d t} \quad \text{with} \quad \vec{p}(t) = m(t) \, \vec{v}(t)\,. $$ Then, for frame A, we have: $$ \begin{align} \vec{F}_C^{(A)} &= \frac{d \vec{p}^{(A)}_C(t)}{d t}\\ & = \frac{d m_C}{dt}\, \vec{v}_{CA} + m_C \, \frac{d \vec{v}_{CA}}{dt} = \frac{d m_C}{dt}\, \vec{v} + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ and for frame B: $$ \begin{align} \vec{F}_C^{(B)} &= \frac{d \vec{p}^{(B)}_C(t)}{d t} \\ & = \frac{d m_C}{dt}\, \vec{v}_{CB} + m_C \, \frac{d \vec{v}_{CB}}{dt} \\ & = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \left(\vec{v} - \vec{u} \right)}{dt} = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ where the last equality follows because $\vec{u}$ is a constant. So we find that: $$ \left(\vec{F}_C^{(A)} - \vec{F}_C^{(B)}\right)(t) = \frac{dm_C(t)}{dt} \, \vec{u} $$ Is that your conclusion?

I think the issue is that your statement of Newton's Second Law is incorrect. The law is the equation of motion for a particle of mass $m$. In that way we can write Newton II in all of these equivalent ways: $$ \vec{F}_\text{net} = m \vec{a} = m \frac{d\vec{v}}{dt} = \frac{d \left( m \vec{v}\right)}{dt} = \frac{d \vec{p}}{dt} $$ exactly because we assume that the particle has constant mass.

If you would like to write down a "Newton's Second Law for an object with changing mass", then you need to derive an effective equation for a system of particles. See, for example, my recent answer here. The result of summing many Newton II equations for many particles is this: $$ \vec{F}_\text{net, external} = M_\text{tot} \frac{d \vec{V}_\text{com}}{d t}\,, $$ where the left-hand side is the net external force on the system (all internal forces have cancelled out via Newton III); $M_\text{tot}$ is the sum of all the particle masses; and the center-of-mass velocity is defined as: $$ \vec{V}_\text{com} = \frac{d}{dt} \vec{R}_\text{com} = \frac{d}{dt} \left( \frac{m_1 \vec{v}_1 + \cdots + m_N \vec{v}_N}{M_\text{tot}} \right)\,. $$

I think you need to frame your question in terms of that extension of Newton's Second Law. I'm not sure how to precisely state such a question, but it would be a nice exercise. I don't think, though, that in doing so you will arrive at such contradictions, because to have a contradiction you'll need to show it happening at the level of the equation of Newton's Second Law for a particle.

In your original statement, which I summarized at the beginning of the answer, there is simply no contradiction because it is impossible, within the theory, for a particle $m$ to have changing mass.

Let me just slightly restate your problem to make sure I'm being clear. Let's say we have frame A and frame B, where $\vec{v}_{BA} = \vec{u}$, meaning the "velocity of B with respect to A" is $\vec{u}$. Then if we have an object, C, with velocity $\vec{v}_{CA} = \vec{v}$, its velocity with respect to frame B is (using the relative velocity equation): $$ \vec{v}_{CB} = \vec{v}_{CA} + \vec{v}_{AB} = \vec{v} + \left( - \vec{v}_{BA} \right) = \vec{v} - \vec{u} \,. $$ Now you are assuming that Newton's Second Law says that the net force on an object over time can be found kinematically from the momentum: $$ F_C = \frac{d \vec{p}_C(t)}{d t} \quad \text{with} \quad \vec{p}(t) = m(t) \, \vec{v}(t)\,. $$ Then, for frame A, we have: $$ \begin{align} \vec{F}_C^{(A)} &= \frac{d \vec{p}^{(A)}_C(t)}{d t}\\ & = \frac{d m_C}{dt}\, \vec{v}_{CA} + m_C \, \frac{d \vec{v}_{CA}}{dt} = \frac{d m_C}{dt}\, \vec{v} + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ and for frame B: $$ \begin{align} \vec{F}_C^{(B)} &= \frac{d \vec{p}^{(B)}_C(t)}{d t} \\ & = \frac{d m_C}{dt}\, \vec{v}_{CB} + m_C \, \frac{d \vec{v}_{CB}}{dt} \\ & = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \left(\vec{v} - \vec{u} \right)}{dt} = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ where the last equality follows because $\vec{u}$ is a constant. So we find that: $$ \left(\vec{F}_C^{(A)} - \vec{F}_C^{(B)}\right)(t) = \frac{dm_C(t)}{dt} \, \vec{u} $$ Is that your conclusion?

I think the issue is that your statement of Newton's Second Law is incorrect. The law is the equation of motion for a particle of mass $m$. In that way we can write Newton II in all of these equivalent ways: $$ \vec{F}_\text{net} = m \vec{a} = m \frac{d\vec{v}}{dt} = \frac{d \left( m \vec{v}\right)}{dt} = \frac{d \vec{p}}{dt} $$ exactly because we assume that particle has constant mass.

If you would like to write down a "Newton's Second Law for an object with changing mass", then you need to derive an effective equation for a system of particles. See, for example, my recent answer here. The result of summing many Newton II equations for many particles is this: $$ \vec{F}_\text{net, external} = M_\text{tot} \frac{d \vec{V}_\text{com}}{d t} $$ where the left-hand side is the net external force on the system (all internal forces have cancelled out via Newton III), $M_\text{tot}$ is the sum of all the particle masses, and the center-of-mass velocity is defined as: $$ \vec{V}_\text{com} = \frac{d}{dt} \vec{R}_\text{com} = \frac{d}{dt} \left( \frac{m_1 \vec{v}_1 + \cdots + m_N \vec{v}_N}{M_\text{tot}} \right)\,. $$ I think you need to frame your question in terms of that equation (I'm not sure how to precisely state it).

But the original statement, which I summarized at the beginning of the answer, is simply not a contradiction because it is impossible, within the theory, for a particle $m$ to have changing mass.

Let me just slightly restate your problem to make sure I'm being clear. Let's say we have frame A and frame B, where $\vec{v}_{BA} = \vec{u}$, meaning the "velocity of B with respect to A" is $\vec{u}$. Then if we have an object, C, with velocity $\vec{v}_{CA} = \vec{v}$, its velocity with respect to frame B is (using the relative velocity equation): $$ \vec{v}_{CB} = \vec{v}_{CA} + \vec{v}_{AB} = \vec{v} + \left( - \vec{v}_{BA} \right) = \vec{v} - \vec{u} \,. $$ Now you are assuming that Newton's Second Law says that the net force on an object over time can be found kinematically from the momentum: $$ F_C = \frac{d \vec{p}_C(t)}{d t} \quad \text{with} \quad \vec{p}(t) = m(t) \, \vec{v}(t)\,. $$ Then, for frame A, we have: $$ \begin{align} \vec{F}_C^{(A)} &= \frac{d \vec{p}^{(A)}_C(t)}{d t}\\ & = \frac{d m_C}{dt}\, \vec{v}_{CA} + m_C \, \frac{d \vec{v}_{CA}}{dt} = \frac{d m_C}{dt}\, \vec{v} + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ and for frame B: $$ \begin{align} \vec{F}_C^{(B)} &= \frac{d \vec{p}^{(B)}_C(t)}{d t} \\ & = \frac{d m_C}{dt}\, \vec{v}_{CB} + m_C \, \frac{d \vec{v}_{CB}}{dt} \\ & = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \left(\vec{v} - \vec{u} \right)}{dt} = \frac{d m_C}{dt} \left(\vec{v} - \vec{u} \right) + m_C \frac{d \vec{v}}{dt}\,, \end{align} $$ where the last equality follows because $\vec{u}$ is a constant. So we find that: $$ \left(\vec{F}_C^{(A)} - \vec{F}_C^{(B)}\right)(t) = \frac{dm_C(t)}{dt} \, \vec{u} $$ Is that your conclusion?

I think the issue is that your statement of Newton's Second Law is incorrect. The law is the equation of motion for a particle of mass $m$. In that way we can write Newton II in all of these equivalent ways: $$ \vec{F}_\text{net} = m \vec{a} = m \frac{d\vec{v}}{dt} = \frac{d \left( m \vec{v}\right)}{dt} = \frac{d \vec{p}}{dt} $$ exactly because we assume that the particle has constant mass.

If you would like to write down a "Newton's Second Law for an object with changing mass", then you need to derive an effective equation for a system of particles. See, for example, my recent answer here. The result of summing many Newton II equations for many particles is this: $$ \vec{F}_\text{net, external} = M_\text{tot} \frac{d \vec{V}_\text{com}}{d t} $$ where the left-hand side is the net external force on the system (all internal forces have cancelled out via Newton III), $M_\text{tot}$ is the sum of all the particle masses, and the center-of-mass velocity is defined as: $$ \vec{V}_\text{com} = \frac{d}{dt} \vec{R}_\text{com} = \frac{d}{dt} \left( \frac{m_1 \vec{v}_1 + \cdots + m_N \vec{v}_N}{M_\text{tot}} \right)\,. $$

I think you need to frame your question in terms of that extension of Newton's Second Law. I'm not sure how to precisely state such a question, but it would be a nice exercise. I don't think, though, that in doing so you will arrive at such contradictions, because to have a contradiction you'll need to show it happening at the level of the equation of Newton's Second Law for a particle.

In your original statement, which I summarized at the beginning of the answer, there is simply no contradiction because it is impossible, within the theory, for a particle $m$ to have changing mass.